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你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用。
示例:
给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
自解1:很傻。。。
class Solution(object):
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
for i in range(0,len(nums)) :
for j in range (0,len(nums)):
if i!=j and nums[i] + nums[j] == target):
return [i,j]
好解法:
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
d={} ## dict字典,是键值对 ##
size=0
while size < len(nums):
if target-nums[size] in d:
if d[target-nums[size]] < size:
return [d[target-nums[size]],size]
else:
d[nums[size]] = size
size += 1 ##### size不加1 死循环 ####